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Spiegel's "Mathematical Handbook of Formulas and Tables" (Schaum, 1968), item $14.655$ gives:

$$\int \frac {\cosh^{-1} (x/a) \, \mathrm d x} {x^2} = \dfrac {-\cosh^{-1} (x/a) } x \mp \dfrac 1 a \ln \left({\dfrac {a + \sqrt {x^2 + a^2} } x}\right)$$

(negative for $\cosh^{-1} (x/a) > 0$, positive for $\cosh^{-1} (x/a) < 0$)

Let's ignore the fact for now that $\cosh^{-1}$ is AFAIK defined as being the $+$ve branch of the function, and roll with what we've got.

It is assumed at this stage that $a > 0$.

Integration by parts:

$u = \cosh^{-1} (x/a) \implies \dfrac {\mathrm d u} {\mathrm d x} = \dfrac 1 {\sqrt {x^2 - a^2} }$

$\dfrac {\mathrm d v} {\mathrm d x} = \dfrac 1 {x^2} \implies v = \dfrac {-1} x$

allows us to assemble:

$$ \int \frac {\cosh^{-1} (x/a) \,\mathrm d x} {x^2} = \frac {-\cosh^{-1} (x/a)} x + \int \frac {\mathrm d x} {x \sqrt {x^2 - a^2} } + C$$

But then we've got this standard integral:

$$\int \frac {\mathrm d x} {x \sqrt {x^2 - a^2} } = \frac 1 a \operatorname {arcsec} \left|{\frac x a}\right| + C$$

In order to return to Spiegel's result, the denominator of the integrand on the RHS of the above would need to be $\displaystyle \int \dfrac {\mathrm d x} {x \sqrt {a^2 - x^2} }$.

But I can't work out how to make it be so. The derivative of $\cosh^{-1} (x/a)$ is pretty well established as being $\dfrac 1 {\sqrt {x^2 - a^2} }$, and you would expect this to be the case, as $\cosh^{-1}$ is defined only on $x > a$ in the first place.

Am I correct in assuming that Spiegel has made a mistake?

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I simply differentiate the second part of Schaum's result and realized their mistake:

for some reason they integrated $$\int\frac{1}{x \sqrt{x^2 + a^2}}\,dx$$ instead of the correct $$\int \frac{1}{x \sqrt {x^2 - a^2} }\,dx$$ Thus the correct integral is $$\int \frac{1}{x^2} \operatorname{arcosh} \left(\frac{x}{a}\right) \,dx = - \frac{1}{x} \operatorname{arcosh} \left(\frac{x}{a}\right) + \frac{1}{a} \operatorname{arcsec} \left(\frac{x}{a}\right) + C $$

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    $\begingroup$ @TravisWillse Thank you! I edited my answer $\endgroup$ – Raffaele Jan 1 at 12:00

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